Question

Solve the equation : $\int \frac{\cos 4x+1}{\cot x-\tan x}dx=?$

Answer 1

$\int \frac{\cos 4x+1}{cotx-tanx}dx=\int \frac{{2\cos }^{2}2x}{\frac{cosx}{sinx}-\frac{sinx}{cosx}}dx[\because \cos 2\theta =2{\cos }^2\theta -1]$

$=2\int \frac{{\cos }^{2}2x.sinx.cosx}{({\cos }^{2}x-{\sin }^{2}x)}dx$

$=\int \frac{{\cos }^{2}2x.2sinxcosx}{{\cos }^{2}x-{\sin }^{2}x}dx$

$=\int \frac{{\cos }^{2}2x.\sin 2x}{\cos 2x}dx[\because {\cos }^{2}x-{\sin }^{2}x=\cos 2x]$

$=\int \cos 2x.\sin 2xdx$ $[\because 2sinx.cosx=\sin 2x]$

$=\frac{1}{2}\int 2\sin 2x.\cos 2xdx$

$=\frac{1}{2}\int \sin 4xdx$

$=-\frac{\cos 4x}{8}+c$